How do you factorise an expression like this? $$u^2v - u^2w - uv^2 + v^2w+uw^2-vw^2.$$
I'm interested in the method on how to get the first factor as I don't see any clear way on how to do this. I don't care so much about the answer as I want to solve it myself. Thanks.
On
Hint. Note that the terms can be rearranged as follows $$(v - w)u^2 + (w-u)v^2 -(v-w+w-u)w^2.$$ Can you take it from here?
On
It's a polynomial of Schur: $$u^2v+v^2w+w^2u-u^2w-v^2u-w^2v=(u-v)(u-w)(v-w).$$
Similarly: $$u^3v+v^3w+w^3u-u^3w-v^3u-w^3v=(u-v)(u-w)(v-w)(u+v+w),$$ $$u^4v+v^4w+w^4u-u^4w-v^4u-w^4v=$$ $$=(u-v)(u-w)(v-w)(u^2+v^2+w^2+uv+uw+vw)...$$
On
We have the expression at hand: $$u^2(v-w)+v^2(u-w)+w^2(u-v)$$
With cyclic coefficients of $u^2, v^2, w^2$, we are motivated to further explore with any one coefficient.
We have, $$(u-v)(w^2)+(u^2v-u^2w+uv^2-v^2w)$$ $$=(u-v)(w^2)+(u^2v-u^2w+uv^2-v^2w+uvw-uvw)$$ $$=(u-v)(w^2)+[(u-v)(-uw)+(u-v)(-vw) +(u-v)(uv)]$$ $$=(u-v)(w^2-uw-v^2+uv)$$ $$=(u-v)(u-w)(v-w)$$
On
Let $$ \eqalign{ A &= u^2v − u^2w − uv^2 + v^2w + uw^2 − vw^2 \cr &= u^2(v − w) − v^2(u - w) + w^2(u − v) \cr } $$ Observe that for $v=w$ then $$ \eqalign{ A &= − w^2(u - w) + w^2(u − w) \cr &= − w^2u - w^3 + w^2u − w^3 \cr &= 0 } $$ thus $(v-w)$ is a factor of $A$. The same we conclude for $(u-w)$ and $(u-v)$, to write finally that $$ A=(v-w)(u-w)(u-v) $$
Note that if $u=v$ the expression reduces to zero, likewise $u=w$ and $v=w$.
The factor theorem for polynomials applies and you can assume factors of $(u-v)$, $(u-w)$ and $(v-w)$. That does most of the work for you (careful about other factors including constants).
It is always worth checking $u=\pm v$ in situations like this in case easy factors drop out.